A hog raiser in a certain province uses two methods of pig-farming: intensive pig farming, where pigs are housed indoors in group-housing or straw-lined sheds; and extensive pig farming, where pigs are allowed to wander around the farm or fence. Test the hypothesis whether or not the mean weight of pigs in intensive farming is better than the extensive farming based from the mean weight of the pigs in the sample with data shown below. Use a one-tailed test at \( \alpha=0.01 \). Intensive farming: \( \bar{X}_{1}=85 \mathrm{~kg}, \mathrm{~s}_{1}{ }^{2}=10 \mathrm{~kg} \), and \( n_{1}=55 \) Extensive farming: \( \bar{X}_{0}=79 \mathrm{~kg}, s_{\eta}{ }^{2}=6 \mathrm{~kg} \), and \( n_{2}=45 \)

**Step 1. State the hypotheses** We are testing whether the mean weight of pigs in intensive pig farming is greater than that in extensive pig farming. Define - \( \mu_1 \) as the true mean weight of pigs in intensive farming, - \( \mu_0 \) as the true mean weight of pigs in extensive farming. The null and alternative hypotheses are \[ H_0: \mu_1 - \mu_0 = 0 \quad \text{vs} \quad H_1: \mu_1 - \mu_0 > 0. \] **Step 2. Given data** For intensive farming: - Sample mean: \( \bar{X}_1 = 85 \, \mathrm{kg} \) - Sample variance: \( s_1^2 = 10 \, \mathrm{kg}^2 \) - Sample size: \( n_1 = 55 \) For extensive farming: - Sample mean: \( \bar{X}_0 = 79 \, \mathrm{kg} \) - Sample variance: \( s_0^2 = 6 \, \mathrm{kg}^2 \) - Sample size: \( n_2 = 45 \) **Step 3. Compute the test statistic** Assuming the two samples are independent and the variances are estimated from the samples, we use the test statistic for the difference of means: \[ t = \frac{(\bar{X}_1 - \bar{X}_0) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_0^2}{n_2}}}. \] Plug in the values: - Difference in means: \[ \bar{X}_1 - \bar{X}_0 = 85 - 79 = 6 \, \mathrm{kg}. \] - Standard error: \[ SE = \sqrt{\frac{10}{55} + \frac{6}{45}}. \] First, compute each term: \[ \frac{10}{55} \approx 0.1818, \quad \frac{6}{45} \approx 0.1333. \] Thus, \[ SE = \sqrt{0.1818 + 0.1333} = \sqrt{0.31515} \approx 0.561. \] Now, compute the t-statistic: \[ t = \frac{6}{0.561} \approx 10.70. \] **Step 4. Determine the degrees of freedom** We use the Welch–Satterthwaite formula: \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_0^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_0^2}{n_2}\right)^2}{n_2-1}}. \] Plug in the numbers: - Numerator is \[ \left(0.1818 + 0.1333\right)^2 = (0.31515)^2 \approx 0.09936. \] - Denominator: For the first term: \[ \frac{(0.1818)^2}{55-1} = \frac{0.03306}{54} \approx 0.000611. \] For the second term: \[ \frac{(0.1333)^2}{45-1} = \frac{0.01778}{44} \approx 0.000404. \] Sum the two parts: \[ 0.000611 + 0.000404 = 0.001015. \] Thus, \[ df \approx \frac{0.09936}{0.001015} \approx 97.9. \] We can use \( df \approx 98 \). **Step 5. Determine the critical value** For a one-tailed test at \( \alpha = 0.01 \) and approximately \( 98 \) degrees of freedom, the critical value \( t_{\alpha, df} \) is approximately \[ t_{0.01, 98} \approx 2.36. \] **Step 6. Compare the test statistic with the critical value** We have \[ t \approx 10.70 \quad \text{and} \quad t_{\text{critical}} \approx 2.36. \] Since \[ 10.70 > 2.36, \] we reject the null hypothesis. **Step 7. Conclusion** At \( \alpha = 0.01 \), there is sufficient evidence to conclude that the mean weight of pigs in intensive farming is higher than that in extensive farming.